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CHE302 Process Dynamics and Control Korea University 9-1

CHE302 LECTURE IX

FREQUENCY RESPONSES

Professor Dae Ryook Yang

Fall 2001

Dept. of Chemical and Biological Engineering Korea University

CHE302 Process Dynamics and Control Korea University 9-2

DEFINITION OF FREQUENCY RESPONSE

• For linear system

– “The ultimate output response of a process for a sinusoidal input at a frequency will show amplitude change and phase shift at the same frequency depending on the process characteristics.”

– Amplitude ratio (AR): attenuation of amplitude, – Phase angle ( ): phase shift compared to input – These two quantities are the function of frequency.

Process

Input Output

sin

A ωt Aˆsin(ω φt+ )

After all transient effects are decayed out.

t φ

( ) u t ( )

y t

-A Aˆ A

Aˆ

φ

ˆ/

A A

CHE302 Process Dynamics and Control Korea University 9-3

BENEFITS OF FREQUENCY RESPONSE

• Frequency responses are the informative representations of dynamic systems

– Audio Speaker

– Equalizer

– Structure

Elec. Signal Sound wave

Raw signal

Processed signal

Expensive speaker

Cheap speaker

Old Tacoma bridge New Tacoma bridge

Old Tacoma bridge

New Tacoma bridge Adjustable

for each frequency

band

CHE302 Process Dynamics and Control Korea University 9-4

– Low-pass filter

– High-pass filter

– In signal processing field, transfer functions are called“filters ”.

(2)

CHE302 Process Dynamics and Control Korea University 9-5

• Any linear dynamical system is completely defined by its frequency response.

– The AR and phase angle define the system completely.

– Bode diagram

• AR in log -log plot

• Phase angle in log -linear plot

– Via efficient numerical technique (fast Fourier transform, FFT), the output can be calculated for any type of input.

• Frequency response representation of a system dynamics is very convenient for designing a feedback controller and analyzing a closed-loop system.

– Bode stability

– Gain margin (GM) and phase margin (PM)

CHE302 Process Dynamics and Control Korea University 9-6

• Critical frequency

– As controller gain changes, the amplitude ratio (AR) and the phase angle (PA) change.

– The frequency where the PA reaches –180° is calledcritical frequency ( ).

– The component of output at the critical frequency will have the exactly same phase as the signal goes through the loop due to comparator (-180 °) and phase shift of the process (-180 °).

– For the open-loop gain at the critical frequency,

• No change in magnitude

• Continuous cycling – For

• Getting bigger in magnitude

• Unstable – For

• Getting smaller in magnitude

• Stable ωc

( ) 1

OL c

K ω =

R Gc… GP

+

-

Y E

Sign change

Sign change

( ) 1

OL c

K ω >

( ) 1

OL c

K ω <

CHE302 Process Dynamics and Control Korea University 9-7

• Example

– If a feed is pumped by a peristaltic pump to a CSTR, will the fluctuation of the feed flow appear in the output?

– V=50cm3, q=90cm3/min (so is the average of qi)

• Process time constant=0.555min.

– The rpm of the peristaltic pump is 60rpm.

• Input frequency=180rad/min (3blades) – The AR=0.01

t qi

V cAi, qi

Peristaltic pump cA, q

If the magnitude of fluctuation of qiis 5% of nominal flow rate, the fluctuation in the output concentration will be about 0.05% which is almost unnoticeable.

( constant)

A

i Ai A

Vdc q c qc q

dt =

/ ( )

( ) ( / ) 1

Ai A i

A i

C C q

C s

q s =Vs q= V q s

+ +

(ωτ =100)

CHE302 Process Dynamics and Control Korea University 9-8

OBTAINING FREQUENCY RESPONSE

• From the transfer function, replace s with

– For a pole, , the response mode is .

– If the modes are not unstable ( ) and enough time elapses, the survived modes becomes . (ultimate response)

• The frequency response, is complex as a function of frequency.

( )

s j

( ) G s 

= ω

G j ω

Transfer function Frequency response

( ) G jω ( ) Re[ ( )] Im[ ( )]

G jω = G jω + j G jω

2 2

( ) Re[ ( )] Im[ ( )]

AR= G jω = G jω + G jω

( )

( ) tan 1 Im[ ( )]/Re[ ( )]

G j G j G j

φ=R ω = ω ω

s= +α e(α+jω)t α≤0

ej tω

Re Im

φ ( ) G jω

AR

ω

Nyquist diagram

Bode plot

(3)

CHE302 Process Dynamics and Control Korea University 9-9

• Getting ultimate response

– For a sinusoidal forcing function – Assume G(s) has stable poles bi.

– Without calculating transient response, the frequency response can be obtained directly from .

– Unstable transfer function does not have a frequency response because a sinusoidal input produces an unstable output response.

2 2

( ) ( ) A Y s G s

s ω

= ω

+

1

2 2 2 2

1

( ) ( ) n

n

A Cs D

Y s G s

s s b s b s

α

ω α ω

ω ω

= = + + + +

+ + L + +

( ) ( ) D C

G j A Cj D G j j R jI

A A

ω ω= ω+ ω ω = + = +

Decayed out at larget

, ( cosul sin ) ˆsin( )

C=IA D=RA y =A I ωt+R ωt = A ωt+φ

2 2 1

ˆ / = ( ) and tan ( / ) ( )

AR A A R I G jω φ I R G jω

∴ = = + = =R

( )

G jω

CHE302 Process Dynamics and Control Korea University 9-10

• First-order process

• Second-order process

( ) ( 1) G s K

τs

= +

2 2

( ) (1 )

(1 ) (1 )

K K

G j j

ω j ωτ

ωτ ω τ

= =

+ +

2 2

( ) 1

N 1 AR G jω

= = ω τ + ( ) tan (1 ) φ=RG jω = − ωτ

( ) 2 2

( 2 1)

G s K

s s

τ ζτ

= + +

2 2

( )

(1 ) 2

G j K ω j

τ ω ζτω

= +

2 2 2 2

( )

(1 ) ( 2 )

A R G jω K

ω τ ζ ω τ

= =

+

1 1

2 2

I m ( ( )) 2

( ) tan t a n

R e ( ( )) 1

G j G j

G j

ω ζωτ

φ ω

ω ω τ

= = = −

R

ζ>1

Corner freq.

CHE302 Process Dynamics and Control Korea University 9-11

• Process Zero (lead)

• Unstable pole

( ) a 1 G s =τ s+

( ) 1 a

G jω = +jωτ

2 2

( ) 1

N a

AR =G jω = +ω τ ( ) tan (1 a) φ=RG jω = ωτ

( ) 1

( 1)

G s = τs

− +

2 2

1 1

( ) (1 )

1 1

G j j

ω j τω

τω τ ω

= = +

+

2 2

( ) 1 1 A R G jω

= = ω τ +

1I m ( ( )) 1

( ) tan tan

R e ( ( ) ) G j G j

G j

φ ω ω ωτ

ω

=R = =

CHE302 Process Dynamics and Control Korea University 9-12

• Integrating process

• Differentiator

• Pure delay process

( ) 1

G s =As G j( ) 1 1 j

jA A

ω = ω= − ω ( ) 1

ARN G j

ω Aω

= =

1 1

( ) tan ( )

0 2

G j π

φ ω

ω

= = = −

R

( ) s G s =eθ

( ) j cos sin

G jω = eθω= θω j θω

( ) 1

A R= G jω =

( ) tan 1tan

φ =RG jω = − θω = −θω ( )

G s =As G j( ω)= jAω ( )

ARN =G jω =Aω

1 1

( ) tan ( )

0 2

G j π

φ ω

ω

= = =

R

(4)

CHE302 Process Dynamics and Control Korea University 9-13

SKETCHING BODE PLOT

• Bode diagram

– AR vs. frequency in log-log plot – PA vs. frequency in semi-log plot – Useful for

• Analysis of the response characteristics

• Stability of the closed-loop systemonly for open-loop stable systems with phase angle curves exhibit a single critical frequency.

1 2 3

( ) ( ) ( ) ( ) ( ) ( ) ( )

a b c

G s G s G s G s =G s G s G s L

L 1 2 3

( ) ( ) ( )

( )

( ) ( ) ( )

a b c

G j G j G j G j G j G j G j

ω ω ω

ω = ω ω ω L

L

1 2 3

( ) ( ) ( )

( )

( ) ( ) ( )

a b c

G j G j G j G j G j G j G j

ω ω ω

ω = ω ω ω L

L

1 2 3

( ) ( ) ( ) ( )

( ) ( ) ( )

a b c

G j G j G j G j

G j G j G j

ω ω ω ω

ω ω ω

= + + +

− − − −

R R R R L

R R R L

CHE302 Process Dynamics and Control Korea University 9-14

• Amplitude Ratio on log-log plot

– Start from steady-state gain at =0. If GOLincludes either integrator or differentiator it starts at or 0.

– Each first-order lag (lead) adds to the slope –1 (+1) starting at the corner frequency.

– Each integrator (differentiator) adds to the slope –1 (+1) starting at zero frequency.

– A delays does not contribute to the AR plot.

• Phase angle on semi-log plot

– Start from 0° or -180° at =0 depending on the sign of steady- state gain.

– Each first-order lag (lead) adds 0 ° to phase angle at =0, adds -90° (+90°) to phase angle at = , and adds -45° (+45°) to phase angle at corner frequency.

– Each integrator (differentiator) adds -90° (+90°) to the phase angle for all frequency.

– A delay adds – to phase angle depending on the frequency.

ω

ω ω

θω

ω

CHE302 Process Dynamics and Control Korea University 9-15

Examples 1. 2.

G s( )=(10s+1)(5Ks+1)(s+1)

5(0.5 1) 0.5

( ) (20 1)(4 1) s e s

G s s s

+

= + +

G1

G2 G3

G4

G5

G1 G2 G3

CHE302 Process Dynamics and Control Korea University 9-16

3. PI: 5. PID:

4. PD:

( ) C 1 1

I

G s K τ s

 

=  + 

 

( ) C 1 1 D

I

G s K s

s τ τ

 

=  + + 

 

Slope=-1 Slope=+1

1/ at 0

Notch I D

ω = τ τ φ= °

1/ at 45

b I

ω = τ φ= − °

Slope=-1

( )

( ) C 1 D G s =K +τ s

Slope=+1

1/ at 45

b D

ω = τ φ= − °

(5)

CHE302 Process Dynamics and Control Korea University 9-17

NYQUIST DIAGRAM

• Alternative representation of frequency response

• Polar plot of ( is implicit)

– Compact (one plot)

– Wider applicability of stability analysis than Bode plot

– High frequency characteristics will be shrunk near the origin.

• Inverse Nyquist diagram: polar plot of

– Combination of different transfer function components is not easy as with Nyquist diagram as with Bode plot.

Re Im

φ ( ) G jω

AR

ω

Nyquist diagram

( ) Re[ ( )] Im[ ( )]

G jω = G jω +j G jω ( )

G jω ω

1 /G j(ω)

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